MHB Solve Quadratic Equation: $\mathcal{E}$ Experiment

[Julio1]

The $\mathcal{E}$ experiment consists in choosing the numbers $a, b$ and $c$. Describe the event $A \,= \,$ the equation $ax^2+bx+c=0$ has two distinct real roots.

Spoiler

Hi !, I have this problem, I understand that the roots of the equation are $x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a},$ i.e., $x_1=\dfrac{-b+\sqrt{b^2-4ac}}{2a}$ y $x_2=\dfrac{-b-\sqrt{b^2-4ac}}{2a},$ but I don't understand that more there that do.

 

[Nono713]

How are the coefficients chosen? The answer is going to depend on that (for instance, if they are integers chosen from a dice throw the probability is going to be computable from basic probability, but if they are complex numbers drawn from a uniform distribution then $A$ is a zero-probability event).

 

[Julio1]

How are the coefficients chosen? The answer is going to depend on that (for instance, if they are integers chosen from a dice throw the probability is going to be computable from basic probability, but if they are complex numbers drawn from a uniform distribution then $A$ is a zero-probability event).

Hello Bacterius, Thanks !

For that the equation $ax^2+bx+c=0$ has real roots, the coefficients $a,b$ and $c$ must be reals. In other hand, if $a,b,c\in \mathbb{C}$ the equation has complex roots, but how solve now the problem?

 

[HallsofIvy]

That wasn't what Bacterius meant when he asked how the coefficients are chosen. What probability distribution is used? It can't be "uniform" because there is no uniform distribution over the entire real numbers.

In any case, the equation ax^2+ bx+ c= 0 has two distinct real roots if and only if the discriminant, b^2- 4ac, is greater than 0. The probability of that depends upon the probability distribution of a, b, and c.